Answer
$\frac{-1}{{2}\sqrt{t}\sqrt( {1}+{t^2})}$
Work Step by Step
Given that $y$ = $cot^{-1}(\sqrt{t})$
on substituting $\sqrt{t}$ = $u$
on applying chain rule derivative
$\frac{dy}{dt}$ = $\frac{{dcot^{-1}{u}}}{du}$ $\times$ $\frac{du}{dt}$
or $\frac{dy}{dt}$ = $-\frac{1}{ (1+{u^2})}$ $\times$ $\frac{du}{dt}$
so $\frac{dy}{dt}$ = $-\frac{1}{ (1+{(\sqrt{t})^2})}$ $\times$ $\frac{d\sqrt {t}}{dt}$
or $\frac{dy}{dt}$ = $-\frac{1}{\sqrt( {1}+{t^2})}$ $\times$ $\frac{1}{2\sqrt{t}}$
The final answer is: $\frac{-1}{{2}\sqrt{t}\sqrt( {1}+{t^2})}$